Q.The displacement of a particle is given by $x = (t - 2)^2$ where $x$ is in metres and $t$ in seconds. The distance covered by the particle in first 4 seconds is

The key idea here is understanding the difference between displacement and distance, especially when a particle changes its direction of motion. To find the total distance covered, we must identify any points in time where the particle reverses its velocity.

First, find the velocity by differentiating the displacement function with respect to time:

$x = (t - 2)^2$

$v = \frac{dx}{dt} = \frac{d}{dt}(t - 2)^2 = 2(t - 2)$

Next, determine if the particle changes direction within the first 4 seconds by setting velocity to zero:

$2(t - 2) = 0 \implies t = 2$ s.

The particle changes direction at $t=2$ s.

Now, calculate the displacement at $t=0$, $t=2$, and $t=4$ seconds:

$x(0) = (0 - 2)^2 = 4$ m

$x(2) = (2 - 2)^2 = 0$ m

$x(4) = (4 - 2)^2 = 4$ m …

The problem asks for the distance covered by the particle, not its displacement. This is a crucial distinction in kinematics.

• Displacement is the net change in position from the starting point to the ending point. It's a vector quantity.
• Distance is the total length of the path traveled, regardless of the direction of motion. It's a scalar quantity and is always non-negative.

If a particle moves in only one direction, the distance covered is simply the magnitude of its displacement. However, if the particle changes its direction of motion during the given time interval, the total distance covered will be greater than the magnitude of its net displacement. To find the total distance, we must identify any points where the particle reverses its direction. This happens when its velocity becomes zero.

Here's how we approach this problem:

1. Determine the particle's initial position.

   The displacement is given by $x = (t - 2)^2$.

   At $t = 0$ s, the initial position is:

   $x(0) = (0 - 2)^2 = (-2)^2 = 4$ m.

   So, the particle starts at $x = 4$ m.

2. Find the velocity function.

   Velocity ($v$) is the rate of change of displacement ($x$) with respect to time ($t$). We find it by differentiating the displacement function:

   Velocity $v = \frac{dx}{dt}$

Using the chain rule, $\frac{d}{dt}(u^n) = n u^{n-1} \frac{du}{dt}$, where $u = (t-2)$:

3. Identify when the particle changes direction. A particle changes its direction of motion when its velocity becomes zero and then changes sign. We set the velocity function to zero to find these turning points:

This means the particle momentarily stops and reverses its direction at $t = 2$ seconds. This time falls within our interval of interest (0 to 4 seconds).

> [!WARNING]
> If you were to simply calculate the magnitude of the net displacement from $t=0$ to $t=4$ s, you would get $|x(4) - x(0)| = |(4-2)^2 - (0-2)^2| = |2^2 - (-2)^2| = |4 - 4| = 0$ m. This is the net displacement, not the total distance covered, because the particle returned to its starting point.

4. Calculate the particle's position at key time points.

We need the position at the start ($t=0$), at the turning point ($t=2$), and at the end of the interval ($t=4$). …

Concept: Reading the Turning Point Straight Off the Vertex Form — No Differentiation Needed

Method: Vertex-Form Recognition + Reflection Symmetry (pure algebra, no calculus at all)

The direct route differentiates $x=(t-2)^2$ to find where the particle reverses. This method instead recognises $(t-2)^2$ as an already-completed square (vertex form) — a parabola's minimum is visible directly from that form, with no derivative required — and then exploits the parabola's own mirror symmetry to get the two half-distances without computing them separately.

Steps

1. Read the vertex directly from the algebraic form. For any expression of the form $(t-a)^2$, the value is a sum of squares $\ge 0$, with equality only at $t=a$ — so $x=(t-2)^2$ has its minimum, $x=0$, exactly at $t=2$, read off without differentiating at all.

2. Verify the curve is symmetric about $t=2$. Substitute $t' = 4-t$ (the reflection of $t$ about the point $t=2$) into $x$:

The two sides are identical — so $x(t)$ is exactly a mirror image of itself about $t=2$, over the range $[0,4]$ (since $t=0$ and $t=4$ are themselves mirror images about $t=2$).

3. Use the symmetry to avoid computing the second half separately. Because the motion from $t=0$ to $t=2$ is the exact time-reversed mirror of the motion from $t=2$ to $t=4$, the distance travelled in the first half must equal the distance travelled in the second half — reflection preserves distances even though it reverses the direction of travel.

4. Compute only the first half's endpoints (no turning-point check needed — Step 1 already located it): …